Lecture 8: Addition, Multiplication & Division Todays topics: - - PowerPoint PPT Presentation

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Lecture 8: Addition, Multiplication & Division

• Today’s topics:
• Signed/Unsigned
• Addition
• Multiplication
• Division

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Signed / Unsigned

• The hardware recognizes two formats:

unsigned (corresponding to the C declaration unsigned int)

• - all numbers are positive, a 1 in the most significant bit

just means it is a really large number signed (C declaration is signed int or just int)

• - numbers can be +/- , a 1 in the MSB means the number

is negative This distinction enables us to represent twice as many numbers when we’re sure that we don’t need negatives

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MIPS Instructions

Consider a comparison instruction: slt $t0, $t1, $zero and $t1 contains the 32-bit number 1111 01…01 What gets stored in $t0?

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MIPS Instructions

Consider a comparison instruction: slt $t0, $t1, $zero and $t1 contains the 32-bit number 1111 01…01 What gets stored in $t0? The result depends on whether $t1 is a signed or unsigned number – the compiler/programmer must track this and accordingly use either slt or sltu slt $t0, $t1, $zero stores 1 in $t0 sltu $t0, $t1, $zero stores 0 in $t0

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Sign Extension

• Occasionally, 16-bit signed numbers must be converted

into 32-bit signed numbers – for example, when doing an add with an immediate operand

• The conversion is simple: take the most significant bit and

use it to fill up the additional bits on the left – known as sign extension So 210 goes from 0000 0000 0000 0010 to 0000 0000 0000 0000 0000 0000 0000 0010 and -210 goes from 1111 1111 1111 1110 to 1111 1111 1111 1111 1111 1111 1111 1110

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Alternative Representations

• The following two (intuitive) representations were discarded

because they required additional conversion steps before arithmetic could be performed on the numbers

• sign-and-magnitude: the most significant bit represents

+/- and the remaining bits express the magnitude

• one’s complement: -x is represented by inverting all

the bits of x Both representations above suffer from two zeroes

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Addition and Subtraction

• Addition is similar to decimal arithmetic
• For subtraction, simply add the negative number – hence,

subtract A-B involves negating B’s bits, adding 1 and A

Source: H&P textbook

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Overflows

• For an unsigned number, overflow happens when the last carry (1)

cannot be accommodated

• For a signed number, overflow happens when the most significant bit

is not the same as every bit to its left

• when the sum of two positive numbers is a negative result
• when the sum of two negative numbers is a positive result
• The sum of a positive and negative number will never overflow
• MIPS allows addu and subu instructions that work with unsigned

integers and never flag an overflow – to detect the overflow, other instructions will have to be executed

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Multiplication Example

Multiplicand 1000ten Multiplier x 1001ten

• 1000

0000 0000 1000

• Product

1001000ten

In every step

• multiplicand is shifted
• next bit of multiplier is examined (also a shifting step)
• if this bit is 1, shifted multiplicand is added to the product

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HW Algorithm 1

In every step

• multiplicand is shifted
• next bit of multiplier is examined (also a shifting step)
• if this bit is 1, shifted multiplicand is added to the product

Source: H&P textbook

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HW Algorithm 2

• 32-bit ALU and multiplicand is untouched
• the sum keeps shifting right
• at every step, number of bits in product + multiplier = 64,

hence, they share a single 64-bit register

Source: H&P textbook

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Notes

• The previous algorithm also works for signed numbers

(negative numbers in 2’s complement form)

• We can also convert negative numbers to positive, multiply

the magnitudes, and convert to negative if signs disagree

• The product of two 32-bit numbers can be a 64-bit number
• - hence, in MIPS, the product is saved in two 32-bit

registers

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MIPS Instructions

mult $s2, $s3 computes the product and stores it in two “internal” registers that can be referred to as hi and lo mfhi $s0 moves the value in hi into $s0 mflo $s1 moves the value in lo into $s1 Similarly for multu

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Fast Algorithm

• The previous algorithm

requires a clock to ensure that the earlier addition has completed before shifting

• This algorithm can quickly set

up most inputs – it then has to wait for the result of each add to propagate down – faster because no clock is involved

• - Note: high transistor cost

Source: H&P textbook

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Division

1001ten Quotient Divisor 1000ten | 1001010ten Dividend

• 1000

10 101 1010

• 1000

10ten Remainder

At every step,

• shift divisor right and compare it with current dividend
• if divisor is larger, shift 0 as the next bit of the quotient
• if divisor is smaller, subtract to get new dividend and shift 1

as the next bit of the quotient

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Division

1001ten Quotient Divisor 1000ten | 1001010ten Dividend 0001001010 0001001010 0000001010 0000001010 100000000000  0001000000 00001000000000001000 Quo: 0 000001 0000010 000001001

At every step,

• shift divisor right and compare it with current dividend
• if divisor is larger, shift 0 as the next bit of the quotient
• if divisor is smaller, subtract to get new dividend and shift 1

as the next bit of the quotient

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Divide Example

• Divide 7ten (0000 0111two) by 2ten (0010two)

Iter Step Quot Divisor Remainder Initial values 1 2 3 4 5

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Divide Example

• Divide 7ten (0000 0111two) by 2ten (0010two)

Iter Step Quot Divisor Remainder Initial values 0000 0010 0000 0000 0111 1 Rem = Rem – Div Rem < 0  +Div, shift 0 into Q Shift Div right 0000 0000 0000 0010 0000 0010 0000 0001 0000 1110 0111 0000 0111 0000 0111 2 Same steps as 1 0000 0000 0000 0001 0000 0001 0000 0000 1000 1111 0111 0000 0111 0000 0111 3 Same steps as 1 0000 0000 0100 0000 0111 4 Rem = Rem – Div Rem >= 0  shift 1 into Q Shift Div right 0000 0001 0001 0000 0100 0000 0100 0000 0010 0000 0011 0000 0011 0000 0011 5 Same steps as 4 0011 0000 0001 0000 0001

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